What will be the work done in blowing out a soap bubble of radius 0.4 cm? The surface tension of soap solution is `30 xx 10^(-3)" N/m".`

Answer

To find the work done in blowing out a soap bubble of radius 0.4 cm, we can follow these steps:

### Step 1: Convert the radius to meters
The radius given is 0.4 cm. We need to convert this to meters for our calculations.

\[ r = 0.4 \, \text{cm} = 0.4 \times 10^{-2} \, \text{m} = 4 \times 10^{-3} \, \text{m} \]

### Step 2: Calculate the total surface area of the soap bubble
A soap bubble has two surfaces (inner and outer). The surface area \(A\) of a sphere is given by the formula:

\[ A = 4\pi r^2 \]

Since there are two surfaces, the total surface area \(A_{total}\) will be:

\[ A_{total} = 2 \times 4\pi r^2 = 8\pi r^2 \]

Substituting the value of \(r\):

\[ A_{total} = 8\pi (4 \times 10^{-3})^2 \]

### Step 3: Calculate \(r^2\)
Calculating \(r^2\):

\[ (4 \times 10^{-3})^2 = 16 \times 10^{-6} = 1.6 \times 10^{-5} \, \text{m}^2 \]

### Step 4: Substitute \(r^2\) into the total surface area formula
Now substituting \(r^2\) back into the total surface area formula:

\[ A_{total} = 8\pi (1.6 \times 10^{-5}) = 12.8\pi \times 10^{-5} \, \text{m}^2 \]

Using \(\pi \approx 3.14\):

\[ A_{total} \approx 12.8 \times 3.14 \times 10^{-5} \approx 40.16 \times 10^{-5} \, \text{m}^2 \]

### Step 5: Calculate the work done using surface tension
The work done \(W\) in creating the bubble is equal to the surface tension \(\gamma\) multiplied by the total surface area \(A_{total}\):

\[ W = \gamma \times A_{total} \]

Given that the surface tension \(\gamma = 30 \times 10^{-3} \, \text{N/m}\):

\[ W = (30 \times 10^{-3}) \times (40.16 \times 10^{-5}) \]

### Step 6: Calculate the work done
Now calculating \(W\):

\[ W = 30 \times 10^{-3} \times 40.16 \times 10^{-5} = 1.2048 \times 10^{-7} \, \text{J} \]

### Step 7: Final Result
Thus, the work done in blowing out the soap bubble is approximately:

\[ W \approx 1.21 \times 10^{-5} \, \text{J} \]